Maths Micro Project.docx

A Micro------project Report On “””””””””””””””””””””””””+++++++++++++++++++++++ ++ models using the concept of differential equations for thermal cooling””””””” Submitted to the MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION, MUMBAI

In partial fulfilment of the requirements for the award of diploma in Mechanical Engineering For the subject of Applied Mathematics Submitted by 1. Sahil kadu 2. Vrushab Jain 3. Sharan Khatri 4. Md salman Ansari 5. Md Parwez alam 6. Abdullaha azam (ME-First Yr. II- Semester) Under the guidance of Guide Name Mr. P.T. Dangore Sr. Lecturer in Mathematics, GHRP, Nagpur

G H RAISONI POLYTECHNIC, NAGPUR Department of Science and Humanities Academic Session 2018-19

CERTIFICATE This is to certify that this micro project report “Prepare modals using the concept of differential equations for thermal cooling” is the bonafide work of Sahil kadu, vrushab jain, Sharan Khatri, Md Salman Ansari, Md parwez alam, Abdullaha Azam (Branch-ME, First Yr. I-Scheme) who carried out project work under my supervision for the partial fulfilment of the requirement for the award of the degree of the diploma in Mechanical Engineering.

Mrs. V. P. Meshram

Mr. P.T. Dangore

First Year coordinator

Project Guide Mr. G. N. Akhade Principal

MAHARASHTRA BOARD OF TECHANICAL EDUCATION, MUMBAI “””””””’’””’”””””’’’

SUBMISSION We, students of G.H. Raisoni Polytechnic, Nagpur, First year (Ischeme) of course MECHANICAL ENGINEERING humbly submit that, we have completed project work as prescribed by Maharashtra state Board of Technical Education, Mumbai, for the subject Applied maths and the project is prescribed in the report by our own skill and study for the academic session 201819, as per the guidance Mr. P.T. Dangore, Lecturer in mathematics , GHRP, Nagpur.

Group Members:

1. Sahil Ghanshyam Kadu 2. Vrushab Dilip jain 3. Sharan Khatri 4. MD Salman Ansari 5. MD parwaz alam 6. Abdullaha azam

Signature of the students:

…………………………………… ….………………………………… …………………………………… …………………………………… …………………………………… ……………………………………

CONTENTS

DE - Modeling - Cooling

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In this section, I will show you some of the examples of building differential equations for cooling & heating. As I mentioned in Governing Equation page, the most important step for cooling/heating case as well is to figure out proper governing equation (governing law). The fundamentals of Cooling problem is based on Newton's Law of Cooling. (I will add some comments on this in my own words later when I have time, but for now I would suggest you to google some basics of Newton's Law of Cooling if you are not familiar with the concept). Cooling with No Temperature input This example can be the most simplest case of Cooling problem and it would show you the essense of Newton's Law of cooling. The situation can be illustrated as shown below. Let's assume that you have a hot drink (like hot coffee) and put it on the table and put a thermometer in it and let it alone for several hours. You would know (from experience) that the temperature will gradually decrease. This example would explain how the temperature would change.

The governing equation for this case can be illustrated as below. Try to follow each and every parts including all the arrows in this illustration and comments, and you would automatically get a differential equation for the situation.

If you just convert the governing law shown above into a matehmatical form, you would get the differential equation as shown below.

Cooling with Temperature input This example is just a little extension to previous example. In this situation, a simple heat source is added. So you would have two factors influencing on the system. One factor is removing heat (cooling) and the other factor is adding heat (heating). The situation can be illustrated as shown below.

The governing equation for this case can be illustrated as below. Try to follow each and every parts including all the arrows in this illustration and comments, and you would automatically get a differential equation for the situation.

If you just convert the governing law shown above into a matehmatical form, you would get the differential equation as shown below.

Home Heating This example would show you a more realistic case but would look much more complicated than the examples shown above. However, don't get scared of it just by the complexity. However complicated a situation look, you can always break the situation into several component which is simple enough for you to attack. Once you get the several simple blocks, just figure out the governing equation for each of the simple block and figure out the differential equation for it. When you complete building the differential equation for all the simpler component blocks, you can simply put all those equations together and get a complete system equation. The situatioin we have to solve is to deduce the mathematical model to represent the temperature over time inside a house. It can be illustrated as shown below. This house is made up of several sectors - Basement, Main Floor,Attic and Surrounding Air. Each of these components has a couple of boundaries with the other components and each of the boundary has different heat tranfer rate.

First, let's deduce the differential equation for the basement. It can be illustrated as shown below. Try to follow each and every parts including all the arrows in this illustration and comments, and you would automatically get a differential equation for the situation.

If you combine all the factors described in the illustration into an equation, you would get a differential equation as shown below.

Next, let's deduce the differential equation for the Main Floor. It can be illustrated as shown below. Try to follow each and every parts including all the arrows in this illustration and comments, and you would automatically get a differential equation for the situation.

If you combine all the factors described in the illustration into an equation, you would get a differential equation as shown below.

First, let's deduce the differential equation for the Attic. It can be illustrated as shown below. Try to follow each and every parts including all the arrows in this illustration and comments, and you would automatically get a differential equation for the situation.

If you combine all the factors described in the illustration into an equation, you would get a differential equation as shown below.

Now if you just put together all the individual equations for each individual component, you would get a system equation (simultaneous equation) as shown below.

The Energy Balance To develop a mathematical model of a thermal system we use the concept of an energy balance. The energy balance equation simply states that at any given location, or node, in a system, the heat into that node is equal to the heat out of the node plus any heat that is stored (heat is stored as increased temperature in thermal capacitances).

Heat in = Heat out + Heat stored To better understand how this works in practice it is useful to consider several examples.

Examples Involving only Thermal Resistance and Capacitance Example: Two thermal resistances in series Consider a situation in which we have an internal temperature, θi, and an ambient temperature, θa with two resistances between them. An example of such a situation is your body. There is a (nearly) constant internal temperature, there is a thermal resistance between your core and your skin (at θs), and there is a thermal resistance between the skin and ambient. We will call the resistance between the internal temperature and the skin temperature Ris, and the temperature between skin and ambient Rsa. a) Draw a thermal model of the system showing all relevant quantities. b) Draw an electrical equivalent c) Develop a mathematical model (i.e., an energy balance). d) Solve for the temperature of the skin if θi, =37°C, θa =9°C, Ris=0.75°/W; for a patch of skin and Rsa= 2.25°/W for that same patch. Solution: a) In this case there are no thermal capacitances or heat sources, just two know temperatures ( θi, and θa), one unknown temperature (θs), and two resistances ( Risand Rsa.)

b) Temperatures are drawn as voltage sources. Ambient temperature is taken to be zero (i.e., a ground "temperature), all other temperatures are measured with respect to this temperature).

c) There is only one unknown temperature (at θs), so we need only one energy balance (and, since there is no capacitance, we don't need the heat stored term).

Note: the first equation included θa, but the second does not, since θa is our reference temperature and is taken to be zero.

d) Solving for θs

Note: you may recognize this result as the voltage divider equation from electrical circuits.

We can now solve numerically (we use 28°C for the internal temperature since it is 28°C above ambient (37°-9°=28°)

This says that θs is 21°C above ambient. Since the ambient temperature is 9°C, the actual skin temperature is 30°C. Note: If Rsa is lowered, for example by the wind blowing, the skin gets cooler, and it feels like it is colder. This is the mechanism responsible for the "wind chill" effect.

Example: Heating a Building with One Room Consider a building with a single room. The resistance of the walls between the room and the ambient is Rra, and the thermal capacitance of the room is Cr, the heat into the room is qi, the temperature of the room is θr, and the external temperature is a constant, θa. a) Draw a thermal model of the system showing all relevant quantities. b) Draw an electrical equivalent c) Develop a mathematical model (i.e., a differential equation). Solution: a) We draw a thermal capacitance to represent the room (and note its temperarature). We also draw a resistance between the capacitance and ambient.

b) To draw the electrical system we need a circuit with a node for the ambient temperature, and a node for the temperature of the room. Heat (a current source) goes into the room. Energy is stored (as an increased temperature) in the thermal capacitance, and heat flows from the room to ambient through the resistor.

c) We only need to develop a single energy balance equation, and that is for the temperature of the thermal capacitance (since there is only one unknown temperature). The heat into the room is qi, heat leaves the room through a resistor and energy is stored (as increased temperature) in the capacitor.

by convention we take the ambient temperature to be zero, so we end up with a first order differential equation for this system.

Example: Heating a Building with One Room, but with Variable External Temperature. Consider the room from the previous example. Repeat parts a, b, and c if the temperature outside is no longer constant but varies. Call the external temperature θe(t) (this will be the temperature relative to the ambient temperature). We will also change the name of the resistance of the walls to Rre to denote the fact that the external temperature is no longer the ambient temperature. Solution: The solution is much like that for the previous example. Exceptions are noted below. a) The image is as before with the external temperature replaced by θe(t).

b) To draw the electrical system we need a circuit with a node for the external temperature and a node for the temperature of the room. Though perhaps not obvious at first we still need a node for the ambient temperature since all of our temperatures are measured relative to this, and our capacitors must always have one node connected to this reference temperature. Heat flows from the room to the external temperature through the resistor.

c) We still only need to develop a single energy balance equation, and that is for the temperature of the thermal capacitance (since there is only one unknown temperature). The heat into the room is qi, heat leaves the room through a resistor and energy is stored (as increased temperature) in the capacitor.

(the ambient temperature is taken to be zero in this equation). In this case we end up with a system with two inputs (qi and θe).

Example: Heating a Building with Two Rooms Consider a building that consists of two adjacent rooms, labeled 1 and 2. The resistance of the walls room 1 and ambient is R1a, between room 2 and ambient is R2a and between room 1 and room 2 is R12. The capacitance of rooms 1 and 2 are C1 and C2, with temperatures θ1 and θ2, respectively. A heater in in room 1 generates a heat qin. The temperaturexternal temperature is a constant, θa. a) Draw a thermal model of the system showing all relevant quantities.

b) Draw an electrical equivalent

c) Develop a mathematical model (i.e., a differential equation). In this case there are two unknown temperatures, θ1and θ2, so we need two energy balance equations. In both cases we will take θa to be zero, so it will not arise in the equations.

Room 1: Heat in = Heat out + Heat Stored

Room 2: Heat in = Heat out + Heat Stored

In this case there are two parts to the "Heat Out" term, the heat flowing through R1a and the heat through R12.

In this case we take heat flow through R12 to (from 1 to 2) to be an input. We could also take this energy balance to have no heat in, and write the heat flow from 2 to 1 as a second "Heat out" term. (note the change of subscripts in the subtracted terms)

The two first order energy balance equations (for room 1 and room 2) could be combined into a single second order differential equation and solved. Details about developing the second order equation are here.

Examples Involving Fluid Flow So far we have not considered fluid flow in any of the examples; let us do so now. Example: Cooling a Block of Metal in a Tank with Fluid Flow. Consider a block of metal (capacitance=Cm, temperature=θm). It is placed in a well mixed tank (at termperature θt, with capacitance Ct). Fluid flows into the tank at temperature θin with mass flow rate Gin, and specific heat cp. The fluid flows out at the same rate There is a thermal resistance to between the metal block and the fluid of the tank, Rmt, and between the tank and the ambient Rta. Write an energy balance for this system.

Note: the resistance between the tank and the metal block, Rmt, is not explicitly shown.

Solution: Since there are two unknown temperatures, we need two energy balance equations.

Metal Block: Heat in = Heat out + Heat Stored

Tank: Heat in = Heat out + Heat Stored

In this case there is not heat in, and heat out is to the tank through Rmt.

In this case we have heat in from the fluid flow and from the metal block. We have heat out to ambient through Rta.

Aside: Modeling a Fluid Flow with and Electrical Analog To model this system with an electrical analog, we can represent the fluid flow as a voltage source at θin, with a resistance equal to 1/(Gin·cp). If you sum currents at the nodes θt and θm you can show that this circuit is equivalent to the thermal system above.